$ A = \left[\begin{array}{rrr}2 & 3 & 4 \\ -2 & -2 & 2\end{array}\right]$ $ E = \left[\begin{array}{rr}-1 & -1 \\ 4 & -2 \\ 4 & 4\end{array}\right]$ What is $ A E$ ?
Because $ A$ has dimensions $(2\times3)$ and $ E$ has dimensions $(3\times2)$ , the answer matrix will have dimensions $(2\times2)$ $ A E = \left[\begin{array}{rrr}{2} & {3} & {4} \\ {-2} & {-2} & {2}\end{array}\right] \left[\begin{array}{rr}{-1} & \color{#DF0030}{-1} \\ {4} & \color{#DF0030}{-2} \\ {4} & \color{#DF0030}{4}\end{array}\right] = \left[\begin{array}{rr}? & ? \\ ? & ?\end{array}\right] $ To find the element at any row $i$ , column $j$ of the answer matrix, multiply the elements in row $i$ of the first matrix, $ A$ , with the corresponding elements in column $j$ of the second matrix, $ E$ , and add the products together. So, to find the element at row 1, column 1 of the answer matrix, multiply the first element in ${\text{row }1}$ of $ A$ with the first element in ${\text{column }1}$ of $ E$ , then multiply the second element in ${\text{row }1}$ of $ A$ with the second element in ${\text{column }1}$ of $ E$ , and so on. Add the products together. $ \left[\begin{array}{rr}{2}\cdot{-1}+{3}\cdot{4}+{4}\cdot{4} & ? \\ ? & ?\end{array}\right] $ Likewise, to find the element at row 2, column 1 of the answer matrix, multiply the elements in ${\text{row }2}$ of $ A$ with the corresponding elements in ${\text{column }1}$ of $ E$ and add the products together. $ \left[\begin{array}{rr}{2}\cdot{-1}+{3}\cdot{4}+{4}\cdot{4} & ? \\ {-2}\cdot{-1}+{-2}\cdot{4}+{2}\cdot{4} & ?\end{array}\right] $ Likewise, to find the element at row 1, column 2 of the answer matrix, multiply the elements in ${\text{row }1}$ of $ A$ with the corresponding elements in $\color{#DF0030}{\text{column }2}$ of $ E$ and add the products together. $ \left[\begin{array}{rr}{2}\cdot{-1}+{3}\cdot{4}+{4}\cdot{4} & {2}\cdot\color{#DF0030}{-1}+{3}\cdot\color{#DF0030}{-2}+{4}\cdot\color{#DF0030}{4} \\ {-2}\cdot{-1}+{-2}\cdot{4}+{2}\cdot{4} & ?\end{array}\right] $ Fill out the rest: $ \left[\begin{array}{rr}{2}\cdot{-1}+{3}\cdot{4}+{4}\cdot{4} & {2}\cdot\color{#DF0030}{-1}+{3}\cdot\color{#DF0030}{-2}+{4}\cdot\color{#DF0030}{4} \\ {-2}\cdot{-1}+{-2}\cdot{4}+{2}\cdot{4} & {-2}\cdot\color{#DF0030}{-1}+{-2}\cdot\color{#DF0030}{-2}+{2}\cdot\color{#DF0030}{4}\end{array}\right] $ After simplifying, we end up with: $ \left[\begin{array}{rr}26 & 8 \\ 2 & 14\end{array}\right] $